JEE PYQ: Laws Of Motion Question 23
Question 23 - 2019 | 10 Apr Shift 1] (MCQ)
A particle of mass m is moving along a trajectory given by $x = x_0 + a\cos\omega_1 t$, $y = y_0 + b\cos\omega_2 t$. The torque, acting on the particle about the origin, at $t = 0$ is :
(1) $m(-x_0 b + y_0 a)\omega_1^2,\hat{k}$
(2) $+my_0 a\omega_1^2,\hat{k}$ [Correct]
(3) zero
(4) $-m(x_0 b\omega_2^2 - y_0 a\omega_1^2)\hat{k}$
Show Answer
Answer: (2)
Solution
At $t = 0$, $\vec{\tau} = +my_0 a\omega_1^2,\hat{k}$.
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